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Hadamard transform on state

This is a little fiddly to get right the first time. Let's start by rewriting eq (3) as $$ \sum_{b\in\{0,1\}}\alpha_b|b\rangle\otimes X^{x_{b,y}}|0\rangle. $$ Now, it probably helps to think of this $...
DaftWullie's user avatar
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What is the intuition behind uniform Hadamard superposition?

The first Hadamard gates applied to a fiducial all-zero's ket on $n$ qubits serve to prepare your input register into the uniform superposition over all $2^n$ basis states. The last Hadamard gates ...
Mark Spinelli's user avatar
5 votes

Show that conjugating $\text{CNOT}$ by $H\otimes H$ exchanges control and target qubits

The key observation is that the CZ gate, whose action on the computational basis is $$ \begin{align} |00\rangle&\mapsto|00\rangle\\ |01\rangle&\mapsto|01\rangle\\ |10\rangle&\mapsto|10\...
Adam Zalcman's user avatar
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How to eliminate the global phase of a state vector?

I think the best way to understand this is via projective geometry. The idea is that in quantum mechanics we always assume that our state vector is normalized and we don't care about global phases, so ...
smitke6's user avatar
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Why isn't $Ry(\pi/2)$ gate equivalent to Hadamard gate?

The algebraic comparison of the operations is, of course, correct. Additionaly, noting that $H$ is Hermitian, we see that $R_y\left(\frac{\pi}{2}\right)$ is not Hermitian, and must be transformed. A ...
inq's user avatar
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Is the plus state a magic state for the Hadamard gate?

Start with a state $|\psi\rangle|+\rangle$. Measure the operator $X_1Z_2$ using either lattice surgery or an ancilla qubit and $CZ,CX$ gates. Call this result $m_{xz}$ Then measure the first qubit in ...
Jahan Claes's user avatar
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How do you write the Hadamard operator on two qubits in braket notation?

Hadamard in Dirac notation is $$ H=\frac{1}{\sqrt{2}}(|0\rangle\langle 0|+|0\rangle\langle 1|+|1\rangle\langle 0|-|1\rangle\langle 1|). $$ For two qubits, you take the tensor product $H\otimes H$. You ...
DaftWullie's user avatar
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4 votes

Mathematical justification of postprocessing in QRNG

A QRNG contains some quantum mechanism that produces a sequence of bits $X_1X_2\dots X_n$ which are then post-processed (using a randomness extractor) to produce some final shorter random string $K_1\...
Rammus's user avatar
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How to express Hadamard gate as a generic trigonometric functions of theta?

In the first case, the gate operates by the mapping $H|1\rangle = \cos\theta|0\rangle - \sin\theta|1\rangle$. The second by $H|1\rangle = -\sin\theta|0\rangle + \cos\theta|1\rangle$. What is the ...
Abdullah Khalid's user avatar
4 votes

Is there a fast sparse Hadamard transform?

This partial answer provides a simple algorithm that generalizes the $k=1$ case and exploits sparsity promise to beat the Walsh-Hadamard transform when $k$ is constant. Specifically, the algorithm ...
Adam Zalcman's user avatar
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Decomposing Hadamard gate

You got the H matrix, up to an unimportant global phase of $\frac{\left(1-i\right)}{\sqrt2}$.
Yaron Jarach's user avatar
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Are equiprobable state same as superposition state

Superposition states of a qubit are usually of form $$| \psi \rangle = a |0\rangle + b |1\rangle $$ where probabilities of getting state $|0\rangle$ and$ |1\rangle$ after a $Z$-measurement are $$p(0) =...
FDGod's user avatar
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How to get all unit vectors of 4 dimension with all entries $\pm 1/2$ using only Hadamard and SWAP transformations

No, you cannot. Consider the state with coefficients $(1,1,1,-1)/2$. This is a maximally entangled state (you can write it as $|0+\rangle+|1-\rangle$). This means that you need an entangling gate to ...
DaftWullie's user avatar
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2 votes

What Hamiltonians generate Hadamard and CNOT matrices?

To add to the other answer: multiple such Hamiltonians are possible, in general. A simple way to see it is to notice that you are looking for Hermitians $H$ such that $e^{iH}=U$ for a given unitary $U$...
glS's user avatar
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In general, what is feasible quantum computation?

I'm guessing that each 'feasible operation' is meant to refer to a specific quantum gate that can be easily implemented on your particular quantum computer of interest, e.g. with one or a couple of ...
Mark Spinelli's user avatar
2 votes

How to eliminate the global phase of a state vector?

Two-level quantum states are typically mapped onto the Bloch sphere in the following way: $$ \left[ \array{c_0\\ c_1} \right] \rightarrow e^{i\alpha} \left[\array{ \cos\frac{\theta}{2} \\ e^{i\phi} \...
jecado's user avatar
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How to eliminate the global phase of a state vector?

You're right, in quantum computing it is customary to discard the global phase of the quantum state, since it can not be observed using measurements. (The probabilities of measurement outcomes are ...
Mariia Mykhailova's user avatar
2 votes

Is the plus state a magic state for the Hadamard gate?

The above circuit shows how to inject a Hadamard gate by means of a magic state $|+\rangle$ and an Ising gate.
Daniele Cuomo's user avatar
2 votes

Inner product in terms of Hadamard and controlled SWAP gates

As mentioned in a comment, this is just a standard swap test, which is asked about repeatedly on this site. It corresponds to a circuit which hopefully shows you more clearly which qubits each step ...
DaftWullie's user avatar
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2 votes

What is the intuition behind uniform Hadamard superposition?

On a basic level, I always thought of it as using Quantum Parallelism. Applying a Hadamard on each input qubit transforms your 'classical' input (e.g. 00) to a quantum input consisting of every ...
K0mp0t1k's user avatar
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How does this measurement in the Hadamard basis look like?

Answer refined based on updated question Your updated question boils down to something like “if we take a Hadamard transform of a superposition of two basis states and measure, why is our string ...
Mark Spinelli's user avatar
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Strange non-normal distribution histogram when applying Hadamard gate in a loop

It is hard to tell exactly what are the arrows pointing at. I assume you are not refering to the vertical statistical disturbance that adds some noise to the expected Gaussian curve (increasing the ...
AG47's user avatar
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Converting $H$ gate to $R_x$ and $R_z$

There's about three little niggles here. First, order of calculation. If you want to get rid of the first $Z$ rotation from your calculation, then that is the right-most one in the product, not the ...
DaftWullie's user avatar
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1 vote
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Hadamard Gate on Cluster States

Your calculation shows that you are doing the right thing for a single step. You just don't seem to be carrying it through for a set of 4 steps. Overall, you should be getting $$ (X^{m_4}HS)(X^{m_3}HS)...
DaftWullie's user avatar
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1 vote

How to eliminate the global phase of a state vector?

This post is also good for understanding the impact of global phase when it comes to using controlled unitary operations: Do global phases matter when a gate is converted into a controlled gate
Quantum Brilliance's user avatar
1 vote

Easy way to look at matrix in computational and Hadamard bases?

If you used a Cholesky decomposition of $M = LL^*$, and then kept $L^*$, then when you need an element in the Hadamard basis, you form $L^*H^{\otimes n}$. From this, just take the relevant columns to ...
Quantum Brilliance's user avatar
1 vote
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How to apply CNOT on a three qubit system, with two qubits already entangled?

The answer to this problem is a combination of (1) The original poster (ME!) had done a tensor product in the wrong order and (2) multiplying by $I \otimes CNOT$ as @jecado suggested. I am providing ...
bddicken's user avatar
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How to eliminate the global phase of a state vector?

Mapping to the Bloch sphere should be global-phase invariant. If you can't cleanly map to the Bloch sphere you essentially skipped a step. For example you can map using $r_i=\langle\psi|\sigma_i|\psi\...
AccidentalTaylorExpansion's user avatar

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